The angle between pair of tangents drawn to the ellipse 3x2+2y2=5 from the point (1, 2) is

A
$\begin{aligned} \tan^{- 1} (12 / 5) \end{aligned}$
B
$\begin{aligned} \tan^{- 1} (6 / \sqrt{5}) \end{aligned}$
C
$\tan^{- 1} (12 / \sqrt{5})$
D
$\tan^{- 1} (8 / \sqrt{5})$

Solution:

$S S_{11} = {S_{1}}^{2}$

$\begin{array}{l} ∴ (3 x^{2} + 2 y^{2} - 5) (3 (1^{2}) + 2 (2^{2}) - 5) = (3 x .1 + 2 y .2 - 5)^{2} \\ \Rightarrow 9 x^{2} - 4 y^{2} - 24 x y + 30 x + 40 y - 55 = 0 \\ \tan θ = 2 \cdot \frac{\sqrt{(h^{2} - a b)}}{a + b}, a = 9, b = - 4, h = - 12 \end{array}$

Alt. The ellipse is of the form $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1 (a^{2} > b^{2})$ the equation of any tangent to it is

$x = m y + \sqrt{a^{2} m^{2} + b^{2}}$ It passes through ( 1, 2l

$∴ (1 - 2 m)^{2} = \frac{5}{6} (3 m^{2} + 2)$

$\Rightarrow$ $9 m^{2} - 24 m - 4 = 0$

$\begin{array}{l} \tan θ = \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} = \frac{{(m_{1} + m_{2})}^{2} - 4 m_{1} m_{2}}{1 + m_{1} m_{2}} \\ = \frac{\sqrt{720}}{5} = \frac{12 \sqrt{5}}{5} = \frac{12}{\sqrt{5}} etc. \end{array}$

The angle between pair of tangents drawn to the ellipse $3 x^{2} + 2 y^{2} = 5$ from the point (1, 2) is

Solution:

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